Problem: $\sum\limits_{n=1}^{\infty } \dfrac{(x-2)^n}{n^2\cdot3^n}$ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-1 < x < 5$ (Choice B) B $0 < x < 4$ (Choice C) C $-1 \le x \le 5$ (Choice D) D $0 \le x \le 4$
We use the ratio test. For $x\neq2$, let $a_n=\dfrac{(x-2)^n}{n^2\cdot3^n}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x-2}{3}\right| $ The series converges when $\left| \dfrac{x-2}{3}\right|<1$, which is when $-1<x<5$. Now let's check the endpoints, $x=-1$ and $x=5$. Letting $x=-1$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(-1-2)^n}{n^2\cdot3^n} &=\sum\limits_{n=1}^{\infty }\dfrac{(-3)^n}{n^2\cdot3^n} \\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{(-1)^n\cdot 3^n}{n^2\cdot3^n} \\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{(-1)^n}{n^2} \end{aligned}$ By the alternating series test, we know this series converges. Letting $x=5$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(5-2)^n}{n^2\cdot3^n} &=\sum\limits_{n=1}^{\infty }\dfrac{3^n}{n^2\cdot3^n} \\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{1}{n^2} \end{aligned}$ By the $p$ -series test, we know this series converges. In conclusion, the interval of convergence is $-1 \le x \le 5$.